Hi I finished this IVP but I cant seem to get the right answer can someone give me some advice as to where I went wrong and point me in the right direction as to how to fix it. Here is the problem and what I have for an answer:
$$y''+9y=\delta(t-2) , y(0)=0, y'(0)=9$$ $$Let Y(s)=L({y(t)})$$ $$s^2Y(s)-sy(0)-y'(0)+9Y(s)=e^{-2s}$$ Plug in Initial conditions
$$Y(s)(s^2+9)=e^{-2s} + 9$$ $$Y(s)=\frac {e^{-2s}}{s^2+9} + \frac{9}{s^2+9}$$ Then using the second shifting theorem we know that: $$L^{-1}(\frac {e^{-2s}}{s^2+9})=u(t-2)f(t-2)$$ Where $$F(s)=\frac{1}{s^2+9} and f(t)=L^{-1}(F(s))=\frac{1}{3}sin(3t)$$ $$Y(s)=u(t-2)(\frac{1}{3}sin(3t-6))+L^{-1}(\frac{9}{s^2+9})$$ Finally What I got was that: $$y(t)=u(t-2)(\frac{1}{3}sin(3t-6)) + 3sin(3t)$$ But the correct answer should be: $$ y(t)=3sin(3t) -\frac{1}{3}u(t-2)sin(6-3t)$$
I am not to sure how to get this correct answer so any suggestions or hints on something I missed would be very much appreciated. Thanks.
You already have the correct answer - since $\sin(-x)=-\sin(x)$, your answer and the "correct" answer are one and the same.